We can easily make a graph of a function over a specified interval. What is not always so easy is to pick an interval that shows off the feature of interest. In the section on rational functions there was a discussion about how to draw graphs for rational functions so that horizontal and vertical asymptotes can be seen. These are properties of the "large." In this section, we build on this but concentrate now on more focused properties: zeros, relative maxima, and relative minima.
Positive and increasing on an interval
Before beginning, we need some vocabulary:
A function $f$ is positive on an interval $I$ if for any $a$ in $I$ it must be that $f(a) > 0$.
Of course, we define negative in a parallel manner. The intermediate value theorem says a continuous function can not change from positive to negative without crossing $0$. This is not the case for functions with jumps, of course.
A function, $f$, is (strictly) increasing on an interval $I$ if for any $a < b$ it must be that $f(a) < f(b)$.
The word strictly is related to the inclusion of the $<$ precluding the possibility of a function being flat for a bit that the $\leq$ inequality would allow.
A parallel definition with $a < b$ implying $f(a) > f(b)$ would be used for a strictly decreasing function.
We can try and prove these properties for a function algebraically – we'll see both are related to the zeros of some function, but, as with most problems, However, before proceeding to that it is usually helpful to get an idea of where the answer is using exploratory graphs.
This helper function plots the function f twice: the second time only when the second function g is positive.
using Plots
plotif(f, g, a, b) = plot([f, x -> g(x) > 0 ? f(x) : NaN], a, b, linewidth=5)
plotif (generic function with 1 method)
To see where a function is positive, we simply pass the function object in for bothf and g above. For example, let's look at where $f(x) = \sin(x)$ is positive:
f(x) = sin(x)
plotif(f, f, -2pi, 2pi)
Let's graph with cos in the masking spot and see what happens:
plotif(sin, cos, -2pi, 2pi)
Maybe surprisingly, we see that the increasing parts of the sine curve are now highlighted. Of course, the cosine is the derivative of the sine function, now we discuss that this is no coincidence.
The derivative and increasing and decreasing
The derivative, $f'(x)$, computes the slope of the tangent line to the graph of $f(x)$ at the point $(x,f(x))$. If the derivative is positive, the tangent line will have an increasing slope. Clearly if we see an increasing function and mentally layer on a tangent, it will have a positive slope. Intuitively then increasing functions and positive derivatives are related concepts. But there are some technicalities. Suppose $f(x)$ has a derivative on $I$ . Then
If $f'(x)$ is positive on an interval $I=(a,b)$, then $f(x)$ is strictly increasing on $I$.
Meanwhile
If a function $f(x)$ is increasing on $I$, then $f'(x) \geq 0$.
The technicality being the equality parts. In the second statement, we have the derivative is non-negative, as we can't guarantee it is positive, even if we considered just strictly increasing functions.
We can see by the example of $f(x) = x^3$ that strictly increasing functions can have a zero derivative, at a point.
The mean value theorem provides the reasoning behind the first statement: on $I$, the slope of any secant line between $d < e$ (both in $I$) is matched by the slope of some tangent line, which by assumption will always be positive. If the secant line slope is written as $(f(e) - f(d))/(e - d)$ with $d < e$, then it is clear then that $f(e) - f(d) > 0$, or $f(e) > f(d)$.
The second part, follows from the secant line equation. The derivative can be written as a limit of secant-line slopes, each of which is positive. The limit of positive things can only be non-negative, though there is no guarantee the limit will be positive.
So, to visualize where a function is increasing, we can just pass in the derivative as the masking function in our plotif function. For example,
using Roots # for D
Base.ctranspose(f::Function) = D(f) # to use f' in place of D(f)
f(x) = sin(pi*x) * (x^3 - 4x^2 + 2)
plotif(f, f', -2, 2)
First derivative test
We know from Fermat that relative maxima and minima occur at critical points. But what is not true is that all critical points correspond to relative maxima and minima. Again, $f(x)=x^3$ provides the example at $x=0$. This is a critical point, but clearly not a relative maximum or minimum – it is just a slight pause for a strictly increasing function.
When will a critical point correspond to a relative maximum? That answer can be answered by considering the first derivative.
The first derivative test: If $c$ is a critical point for $f(x)$ and if $f'(x)$ changes sign at $x=c$, then $f(c)$ will be either a relative maximum or a relative minimum. It will be a maximum if the derivative changes sign from $+$ to $-$ and a minimum if the derivative changes sign from $-$ to $+$. If $f'(x)$ does not change sign at $c$, then $(c,f(c))$ is not a relative maximum or minimum.
The classification part, should be clear: if the derivative is positive then negative, the function $f$ will increase to $(c,f(c))$ then decrease from $(c,f(c))$ and vice versa.
Our definition of critical point assumes $f(c)$ exists, as $c$ is in the domain of $f$. With this assumption, vertical asymptotes are avoided. However, it need not be that $f'(c)$ exists. The absolute value function at $x=0$ provides an example: this point is a critical point where the derivative changes sign, but is not defined exactly at $x=0$. Regardless, it is guaranteed that $(c,f(c))$ will be a relative minimum by the first derivative test.
Example
Consider the function $f(x) = e^{-\lvert x\rvert} \cos(\pi x)$ over $[-3,3]$:
We can see the first derivative test in action – at the peaks and valleys, the relative extrema, the color changes, as the function changes from increasing to decreasing or vice versa.
Example
Find all the relative maxima and minima of the function $f(x) = \sin(\pi \cdot x) \cdot (x^3 - 4x^2 + 2)$ over the interval $[-2, 2]$.
We will do so numerically, rather than attempt this algebraically. For this task we first need to gather the critical points. As each of the pieces of $f$ are everywhere differentiable and no quotients are involved, the function $f$ will be everywhere differentiable. As such, only zeros of $f'(x)$ can be critical points. We find these with
We should be careful though, as fzeros may miss zeros that are not simple or too close together. A critical point will be a simple zero if the function crosses the axis, so these can not be "pauses." As this is exactly the case we are screening for, we double check that all the critical points are accounted for by graphing the derivative:
plot([f', zero], -2, 2)
We see the five zeros as stored in cps and note that at each the function clearly crosses the $x$ axis.
From this last graph of the derivative we can also characterize the graph of $f$: The left-most critical point coincides with a relative minimum of $f$, as the derivative changes sign from negative to positive. The critical points then alternate relative maximum, relative minimum, relative maximum, and finally relative minimum.
Example
Consider the function $f(x) = \sqrt{\lvert x^2 - 1\rvert}$. Find the critical points and characterize them as relative extrema or not.
We will apply the same approach, but need to get a handle on how large the values can be. The function is a composition of three functions. We should expect that the only critical points will occur when the interior polynomial, $x^2-1$ has values of interest, which is around $-1$ to $1$. So we look on the interval $[-2, 2]$:
f(x) = sqrt(abs(x^2 - 1))
cps = fzeros(f', -2, 2)
3-element Array{Float64,1}:
-1.0
-0.0
1.0
We see the three values $-1$, $0$, $1$ that correspond to the two zeros and the relative minimum of $x^2 - 1$. We could graph things, but instead we characterize these values using a sign chart. A continuous function only can change sign when it crosses $0$ and the derivative will be continuous, except possibly at the three values above.
We can then pick intermediate values to test for positive or negative values:
test_pts = [-2, -1/2, 1/2, 2]
[f'(x) for x in test_pts]
the derivative changes sign from negative to postive at $x=-1$, so $f(x)$ will have a relative minimum.
the derivative changes sign from positive to negative at $x=0$, so $f(x)$ will have a relative maximum.
the derivative changes sign from negative to postive at $x=1$, so $f(x)$ will have a relative minimum.
We did this all without graphs. But, let's look at the graph of the derivative:
plot(f', -2, 2)
We see asymptotes at $x=-1$ and $x=1$! These aren't zeroes of $f'(x)$, but rather where $f'(x)$ does not exist. The conclusion is correct – each of $-1$, $0$ and $1$ are critical points – but not for the same reason.
plot(f, -2, 2)
Finally, why does fzeros find these values that are not zeros of $f'(x)$? It uses the bisection algorithm on bracketing intervals to find zeros which are guaranteed by the intermediate value theorem. But we see from the graph that $f'(x)$ is not continuous. Still the algorithm will also converge to values where the function jumps over $0$, which this function clearly does.
Example
Consider the function $f(x) = \sin(x) - x$. Characterize the critical points.
We will work symbolically for this example.
using SymPy
x = symbols("x", real=true)
f(x) = sin(x) - x
cps = solve(diff(f(x)), x)
\begin{bmatrix}0\\2 \pi\end{bmatrix}
We get values of $0$ and $2\pi$. Let's look at the derivative at these points:
At $x=0$ we have to the left and right signs found by
[subs(diff(f), x=>y) for y in [-1/10, 1/10]]
2-element Array{Any,1}:
cos(x) - 1
cos(x) - 1
Both are negative. The derivative does not change sign at $0$, so the critical point is neither a relative minimum or maximum.
What about at $2\pi$? We do something similar:
[subs(diff(f), x=>y) for y in 2*pi + [-1/10, 1/10]]
2-element Array{Any,1}:
cos(x) - 1
cos(x) - 1
Again, both negative. The function $f(x)$ is just decreasing near $2\pi$, so again the critical point is neither a relative minimum or maximum.
A graph verifies this:
plot(f, -3pi, 3pi)
We see that at $0$ and $2\pi$ there are "pauses" as the function decreases. We should also see that this pattern repeats. The critical points found by solve are only those within a certain domain. Any value that satisfies $\cos(x) - 1 = 0$ will be a critical point, and there are infinitely many of these of the form $k \cdot 2\pi$ for $k$ an integer.
We can see the slant asymptote and hints of vertical asymptotes, but, we'd like to see much more of the basic features of the graph.
Previously, we have discussed rational functions and their asymptotes. This function has numerator of degree 3 and denominator of degree 2, so will have a slant asymptote. As well, the zeros of the denominator, $0$ and $-2$, will lead to vertical asymptotes. To get all those into one graph, we used a function to trim the large $y$ values:
trim(f, M=10) = x -> abs(f(x)) > M ? NaN : f(x)
plot(trim(f), -10, 10)
In addition, we want to make sure we include the zeros of $f(x)$. These are the zeros of the numerator, or $1$ and $3$.
As well, we want to make sure the relative minima are shown. These are the critical points of $f$, and are found with:
cps = fzeros(f', -4, 4)
2-element Array{Float64,1}:
-2.15276
3.0
The values of $f$ at the critical points are:
[f(cp) for cp in cps]
2-element Array{Any,1}:
-9.3635
0.0
Its $0$. In general we don't want to trim these values off when plotting, so we will use 20, not the default of 10, when trimming.
We check that $(0,f(0))$ is a relative minimum, by checking values of the derivative to the left and right:
[sign(f'(x)) for x in [2.5, 3.5]]
2-element Array{Any,1}:
-1.0
1.0
We see the pattern negative to positive that indicates a minimum.
Similarly, we can see that the other critical point is a relative maximum:
[sign(f'(x)) for x in cps[1] + [-.1, .1]]
2-element Array{Any,1}:
1.0
-1.0
At this point, we make our final graph, adjusting the $x$-range to show some of the slant asymptote:
plot(trim(f, 20), -5, 7)
Getting more of the vertical asymptotes shown can be done with lower-level plotting, but this clearly shows the key features of interest and requires not much more than identifying where the points of interest are so we can see them clearly.
Concavity
Consider the function $f(x) = x^2$. Over this function we draw some secant lines:
The graph attempts to illustrate that for this function the secant line between any two points $a$ and $b$ will lie above the graph.
This is a special property not shared by all functions.
Concave up: A function $f(x)$ is concave up on $I$ if for any $a < b$ in $I$ the secant line between $a$ and $b$ lies above the graph of $f(x)$ over $[a,b]$.
A similar definition exists for concave down where the secant lines lie below the graph. Notationally, concave up says $f(a) + (f(b) - f(a))/(b-a) \cdot (x-a) \geq f(x)$ for any $x$ in $[a,b]$. Replacine $\geq$ with $leq$ defines concave up, and with either $>$ or $<$ will add the prefix "strictly." These definitions are useful for a general definition of convex functions. We won't work with these definitions, rather we will characterize concavity for functions which have either a first or second derivative.
If $f'(x)$ exists and is increasing on $(a,b)$ then $f(x)$ is concave up on $(a,b)$.
A proof of this makes use of the same trick used to establish the mean value theorem from Rolle's theorem. Let $g(x) = f(x) - (f(a) + M \cdot (x-a)$, where $M$ is the slope of the secant line between $a$ and $b$. If $f'(x)$ is increasing, then so is $g'(x) = f'(x) + M$. Concave up means $g(x) \leq 0$. Suppose that there is a value where $g(x) > 0$ in $[a,b]$. Then assuming $g'(x)$ always exists, after some work, Rolle's theorem will ensure there is a value where $g'(c) = 0$ and $(c,g(c))$ is a relative maximum. The first derivative test then ensures that $g'(x)$ will increase to the left of $c$ and decrease to the right of $c$. But this can't happen as $g'(x)$ is assumed to be increasing on the interval.
Similarly, if a function has a decreasing derivative on $[a,b]$ then it will be concave down on $[a,b]$.
The relationship between increasing functions and their derivatives gives this second characterization of concavity when the second derivative exists:
If $f''(x)$ exists and is positive on $(a,b)$, then $f(x)$ is concave up on $(a,b)$.
This follows, as we can think of $f''(x)$ as just the first derivative of the function $f'(x)$, so the assumption will force $f'(x)$ to be increasing, and hence $f(x)$ to be concave up.
Example
Let's look at the function $x^2 \cdot e^{-x}$ for positive $x$. A quick graph shows the function is concave up, then down, then up in the region plotted:
f(x) = x^2 * exp(-x)
plotif(f, f'', 0, 8)
From the graph, we would expect that the second derivative – which is continuous – would have two zeros on $[0,8]$:
ips = fzeros(f'', 0, 8)
2-element Array{Float64,1}:
0.585786
3.41421
As well, between the zeros we should have the sign pattern +, -, and +, as we verify:
[sign(f''(x)) for x in [0,1,4]]
3-element Array{Any,1}:
1.0
-1.0
1.0
Second derivative test
Concave up functions are "opening" up, and often just $U$-shaped. At a relative minimum, the graph will be concave up, and conversely concave down at a relative maximum. This observation becomes:
The second derivative test: If $c$ is a critical point of $f(x)$ with $f''(c)$ existing in a neighborhood of $c$, then $(c,f(c))$ will be a relative maximum if $f''(c) > 0$ and a relative minimum if $f''(c) < 0$.
If $f''(c)$ is positive in an interval about $c$, then $f''(c) > 0$ implies the function is concave up at $x=c$, Concave up implies the derivative is increasing so must go from negative to positive at the critical point.
The second derivative test is inconclusive when $f''(c)=0$. No such general statement exists, as there isn't enough information. For example, the function $f(x) = x^3$ has $0$ as a critical point, $f''(0)=0$ and the value is not a relative maximum or minimum. On the other hand $f(x)=x^4$ has $0$ as a critical point, $f''(0)=0$ and is a relative minimum.
Example
Use the second derivative test to characterize the critical points of $f(x) = x^5 - x^4 + x^3$.
f(x) = x^5 - 2x^4 + x^3
cps = fzeros(f', -3, 3)
3-element Array{Float64,1}:
1.97517e-8
0.6
1.0
We can check the sign of the second derivative for each critical point:
[f''(cp) for cp in cps]
3-element Array{Any,1}:
1.1851e-7
-0.72
2.0
That $f''(0.6) < 0$ implies that at $0.6$, $f(x)$ will have a relative maximum. As $f''(1) > 0$, the second derivative test says at $x=1$ there will be a relative minimum. That $f''(0) = 0$ says that only that there may be a relative maximum or minimum at $x=0$, as the second derivative test does not speak to this situation.
This should be consistent with this graph, where $-0.25$, and $1.25$ are chosen to capture the zero at $0$ and the two relative extrema:
plotif(f, f'', -0.25, 1.25)
For the graph we see that $0$ is not a relative maximum or minimum. We could have seen this numerically by checking the first derivative test, and noting there is no sign change:
[sign(f'(x)) for x in [-0.1, 0.1]]
2-element Array{Any,1}:
1.0
1.0
Inflection points
An inflection point is a value where the second derivative of $f$ changes sign. At an inflection point the derivative will change from increasing to decreasing (or vice versa) and the function will change from concave up to down (or vice versa).
We can use the fzeros function to find inflection points, by passing in the second derivative function. For example, consider the bell-shaped function
$$~
f(x) = e^{-x^2/2}.
~$$
A graph suggests relative a maximum at $x=0$, a horizontal asymptote of $y=0$, and two inflection points:
f(x) = exp(-x^2/2)
plotif(f, f'', -3, 3)
The inflection points can be found directly, if desired, or numerically with:
fzeros(f'', -3, 3)
2-element Array{Float64,1}:
-1.0
1.0
Example
A car travels from a stop for 1 mile in 2 minutes. A graph of its position as a function of time might look like any of these graphs:
All three graphs have the same average velocity which is just the $1/2$ miles per minute (30 miles an hour). But the instantaneous velocity – which is given by the derivative of the position function) varies.
The graph f1 has constant velocity, so the position is a straight line with slope $v_0$. The graph f2 is similar, though for first and last 30 seconds, the car does not move, so must move faster during the time it moves. A more realistic graph would be f3. The position increases continuously, as do the others, but the velocity changes more gradually. The initial velocity is less than $v_0$, but eventually gets to be more than $v_0$, then velocity starts to increase less. At no point is the velocity not increasing, for f3, the way it is for f2 after a minute and a half.
The rate of change of the velocity is the acceleration. For f1 this is zero, for f2 it is zero as well – when it is defined. However, for f3 we see the increase in velocity is positive in the first minute, but negative in the second minute. This fact relates to the concavity of the graph. As acceleration is the derivative of velocity, it is the second derivative of position – the graph we see. Where the acceleration is positive, the position graph will be concave up, where the acceleration is negative the graph will be concave down. The point $t=1$ is an inflection point, and would be felt by most riders.
Questions
Question
Consider this graph:
On what intervals (roughly) is the function positive?
Question
Consider this graph:
On what intervals (roughly) is the function negative?
Question
Consider this graph
On what interval(s) is this function increasing?
Question
Consider this graph
On what interval(s) is this function concave up?
Question
Consider this graph
What kind of asymptotes does it appear to have?
Question
If it is known that:
A function $f(x)$ has critical points at $x=-1, 0, 1$
$f'(-2) = 1$ and $f'(-1/2) = -1$
What can be concluded?
Question
Mystery function $f(x)$ has $f'(2) = 0$ and $f''(0) = 2$. What is the most you can say about $x=2$?
Question
Find the smallest critical point of $f(x) = x^3 e^{-x}$.
Question
How many critical points does $f(x) = x^5 - x + 1$ have?
Question
How many inflection points does f(x) = x^5 - x + 1 $ have?
Question
At $c$, $f'(c) = 0$ and $f''(c) = 1 + c^2$. Is $(c,f(c))$ a relative maximum? ($f$ is a "nice" function.)
Question
At $c$, $f'(c) = 0$ and $f''(c) = c^2$. Is $(c,f(c))$ a relative maximum? ($f$ is a "nice" function.)
Question
The graph shows $f'(x)$. Is it possible that $f(x) = e^{-x} \sin(\pi x)$?
Question
The graph shows $f'(x)$. Is it possible that $f(x) = x^4 - 3x^3 - 2x + 4$?
Question
The graph shows $f''(x)$. Is it possible that $f(x) = (1+x)^{-2}$?
Question
This plot shows the graph of $f'(x)$. What is a possible function for $f(x)$?